Plugging in for the Earth's mass and for G, we have 11.2 km/s for the escape velocity for an object launched from the Earth's surface. This is about 25,000 miles per hour
Well YOU cant make your questions verified. if theres more than one answer to the question the person who asked the question will choose the best answer. usually based upon like how well it explains and if its the right answer
Answer:
its an antique physics apparatus for demonstrating acoustic standing waves in a tube.
D is the correct answer!!
Your question kind of petered out there towards the end and you didn't specify
the terms, so I'll pick my own.
The "Hubble Constant" hasn't yet been pinned down precisely, so let's pick a
round number that's in the neighborhood of the last 20 years of measurements:
<em>70 km per second per megaparsec</em>.
We'll also need to know that 1 parsec = about 3.262 light years.
So the speed of your receding galaxy is
(Distance in LY) x (1 megaparsec / 3,262,000 LY) x (70 km/sec-mpsc) =
(150 million) x (1 / 3,262,000) x (70 km/sec) =
<em>3,219 km/sec </em>in the direction away from us (rounded)
